The gravitational attraction between an object and the Earth

Answer:

B

Explanation:

There are a pair of forces acting on the 2 interacting objects

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Answer:

in ine experiment the teacher determined

the initial velocity of object B was 1.475 meters per second.

A conservation law states that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.

Given, A is a sphere that is traveling with a velocity of (3.7) m/s and had a mass of 5 kg.

It collides with sphere B and both particles move together with a velocity of (1.4) m/s after the collision.

From the law of conservation:

Momentum will be the same in case of collision

p₁ = p₂

m₁ v₁ = m₂ v₂

Where P₁ = initial momentum of the system,

P₂ = final momentum of the system,

m₁ = mass of the first object,

v₁ = velocity of the first object,

m₂ = mass of the second object

v₂ = velocity of the second object.

In our case,

m₁ = 5 kg

v₁ = 3.7

Since the mass of B is given 4 kg

m₂ = 4 kg

The final mass will be M = m₁ + m₂

M = 4 + 5

M = 9 Kg

Final velocity V = 1.4 m/s

Applying conservation of momentum:

m₁ v₁ + m₂ v₂ = M V

5 * 3.7 + 4 v = 9 * 1.4

18.5 + 4v = 12.6

4v = 18.5 - 12.6

4v = 5.9

v = 1.475

thus, the initial velocity of the object B was 1.475 meters per second.

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Which statement is true of the electromagnetic spectrum? The correct statement is: the higher the radiant energy, the higher the frequency, and the shorter the wavelength.

In the electromagnetic spectrum, energy, frequency, and wavelength are all related. As the radiant energy increases, the frequency also increases, and the wavelength becomes shorter.

Conversely, when the radiant energy decreases, the frequency becomes lower, and the wavelength becomes longer. This relationship is represented by the formula:

Energy (E) = Planck's constant (h) x Frequency (ν)

and

Frequency (ν) = Speed of light (c) / Wavelength (λ)

By understanding these relationships, it is clear that the correct statement is the one that states that higher radiant energy corresponds to higher frequency and shorter wavelengths in the electromagnetic spectrum.

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f' is a shortest track from S to T that consists of line segments only.

Theorem 2 states that the distance between points s and t on a cylindrical surface is equal to the length of the shortest track in the strip m0 m1. This track f consists of curves f1,f2 ,…,fn, where f1 starts at point S covering s, fn ends at point T covering t, and for each i=1,2,…,n−1, fi+1 starts at the point opposite the endpoint of its predecessor fi. An inhabitant of the strip can move about the strip with unit speed, and make free use of the jet service. The distance in Σ between s and t is equal to the minimum time needed to travel from S to T.

In order to prove that in the definition of distance between points of Σ given in Theorem 2, it is sufficient to consider only tracks f for which each curve fi is a line segment, we proceed as follows:

Proof:Let f be a shortest track in the strip m0 m1, consisting of curves f1,f2 ,…,fn. We need to show that there exists a track f' consisting of line segments only, such that f' is a shortest track from S to T. Consider the curves fi, i = 1, 2, ..., n - 1, which are not line segments. Each such curve can be approximated arbitrarily closely by a polygonal path consisting of line segments. Let f'i be the polygonal path that approximates fi. Then, we have:f' = (f1, f'2, f'3, ..., f'n)where f'1 = f1, f'n = fn, and f'i, i = 2, 3, ..., n - 1, is a polygonal path consisting of line segments that approximates fi.Let l(f) and l(f') be the lengths of tracks f and f', respectively. By the triangle inequality and the fact that the length of a polygonal path is the sum of the lengths of its segments, we have:l(f') ≤ l(f1) + l(f'2) + l(f'3) + ... + l(f'n) ≤ l(f)

Therefore, f' is a shortest track from S to T that consists of line segments only.

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Answer:

Assume that \(g = 9.81\; {\rm m\cdot s^{-2}}\). Also assume that that the air resistance on this stone is negligible.

The total air time of this stone is approximately \(1.5\; {\rm s}\).

The maximum height of this stone is \(1.1\; {\rm m}\), same as the height of the slingshot.

The range of this stone is approximately \(23\; {\rm m}\).

The final velocity of this stone is approximately \(21\; {\rm m \cdot s^{-1}}\).

Explanation:

If the air resistance on this stone is negligible, this stone will accelerate towards the ground with a vertical acceleration \(a_{y}\) of \(a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}\).

At the same time, the horizontal velocity \(v_{x}\) of this stone will stay the same during the entire flight: \(v_{x} = 15.2\; {\rm m\cdot s^{-1}}\).

The stone was launched from a height of \(1.1\; {\rm m}\) above the ground. Therefore, the vertical displacement \(x_{y}\) of the stone will be \(x_{y} = (-1.1)\; {\rm m}\) when the stone hits the ground.

The initial vertical velocity \(u_{y}\) of this stone would be \(u_{y} = 0\; {\rm m\cdot s^{-1}}\) since the stone was launched horizontally.

Let \(v_{y}\) denote the final vertical velocity of this stone right before landing. Apply the SUVAT equation \((v_{y})^{2} - (u_{y})^{2} = 2\, a\, x\) to find \(v_{y}\!\):

\((v_{y})^{2} - (u_{y})^{2} = 2\, a\, x\).

\((v_{y})^{2} = (u_{y})^{2} + 2\, a\, x\).

\(\begin{aligned}v_{y} &= -\sqrt{(u_{y})^{2} + 2\, a\, x} \\ &= -\sqrt{0^{2} + 2\, (-9.81) \, (-1.1)}\; {\rm m\cdot s^{-1}} \\ &\approx -14.6908 \end{aligned}\).

(Final vertical velocity \(v_{y}\) is negative since the stone is travelling downwards toward the ground.)

Let \(t\) denote the duration of this flight. Apply the SUVAT equation \(t = (v_{y} - u_{y}) / (a)\) to find \(t\!\):

\(\begin{aligned} t &= \frac{v_{y} - u_{y}}{a} \\ &\approx \frac{(-14.6908) - 0}{(-9.81)}\; {\rm s} \\ &\approx 1.49753\; {\rm s} \end{aligned}\).

In other words, this stone was in the air for approximately \(1.5\; {\rm s}\).

Also because the stone was launched horizontally, the vertical velocity of this stone started at \(u_{y} = 0\; {\rm m\cdot s^{-1}}\) and became negative (toward the ground) immediately after.

Hence, this stone would not have travelled upward during the entire flight. The height of this stone would be maximized immediately after the stone was launched: \(1.1\; {\rm m}\).

Multiple the horizontal velocity \(v_{x} = 15.2\; {\rm m\cdot s^{-1}}\) of this stone by the duration of the flight \(t \approx 1.49753\; {\rm s}\) to find the range (horizontal displacement) of this stone:

\(\begin{aligned} (\text{range}) &= v_{x}\, t \\ &\approx (15.2\; {\rm m\cdot s^{-1}})\, (1.49753\; {\rm s}) \\ &\approx 23\; {\rm m} \end{aligned}\).

Right before landing, the stone would be travelling with a vertical velocity of \(v_{y} \approx (-14.6908)\; {\rm m\cdot s^{-1}}\) and a horizontal velocity of \(v_{x} = 15.2\; {\rm m\cdot s^{-1}}\). Apply the Pythagorean Theorem to find the overall velocity of this stone at that moment:

\(\begin{aligned} v &= \sqrt{{(v_{x})}^{2}\, {(v_{y})^{2}} \\ &\approx \sqrt{(15.2\; {\rm m\cdot s^{-1})^{2} + (-14.6908)^{2}}\; {\rm m\cdot s^{-1}} \\ &\approx 21\; {\rm m\cdot s^{-1}} \end{aligned}\).

A stack of one million quarters would reach a height of approximately 1.75 kilometers (1.086 miles).

Assuming each quarter has a thickness of 1.75 mm, one million quarters would have a total height of:

1,000,000 quarters x 1.75 mm/quarter = 1,750,000 mm

To convert to kilometers, we divide by 1,000,000:

1,750,000 mm / 1,000,000 = 1.75 km

To convert to miles, we divide by 1.609:

1.75 km / 1.609 = 1.086 miles

Therefore, a stack of one million quarters would reach a height of approximately 1.75 kilometers (1.086 miles).

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1. The height of the image is 0.287 m.

2. The wavelength of the light is 563 nm.

1. The image distance, denoted as `i`, is determined by the lens formula: `1/f = 1/o + 1/i`, where `f` represents the focal length, `o` is the object distance, and `i` represents the image distance. Given `f = 28.3 cm` and `o = 1.78 m`, we need to convert the object distance from meters to centimeters: `o = 1.78 m = 178 cm`. Therefore, the image distance is calculated as follows:

i = (1/f - 1/o)^-1 = (1/28.3 - 1/178)^-1 = 24.53 cm.

The image height, denoted as `h'`, can be determined using the object height `h` and the magnification `m` relationship: `h' = m * h`. The magnification `m` is given by `m = -i/o`, where the negative sign indicates an inverted image. Thus,

m = -i/o = -(24.53 cm)/(178 cm) = -0.138.

The image height `h'` is obtained by multiplying `h` by `m`: `h' = m * h`, where `h = 2.08 m`. Therefore,

h' = (-0.138) * 2.08 = -0.287 m.

The negative sign signifies an inverted image. Hence, the height of the image is determined as `0.287 m`, and it is inverted.

2. Bright fringes are observed at angles `theta` satisfying the condition `d sin theta = m lambda`, where `d` represents the spacing between two slits, `m` is an integer indicating the fringe order, and `lambda` denotes the wavelength of light. In this case, given `d = 1/20 mm` and `m = 1`, the angle `theta` corresponding to the first bright fringe is given by `tan theta = x/L`, where `x` represents the separation between two fringes, and `L` is the distance from the grating to the screen. With `x = 27.2 mm` and `L = 2.41 m`, we can calculate:

tan theta = (27.2 mm)/(2.41 m) = 0.01126.

Therefore, `sin theta = tan theta = 0.01126`.

Consequently, the wavelength `lambda` is determined using the formula `lambda = d sin theta / m`, where `d = 1/20 x 10^-3 m`, `sin theta = 0.01126`, and `m = 1`:

lambda = (1/20 x 10^-3 m) x 0.01126 / 1 = 5.63 x 10^-7 m = 563 nm.

In summary:

1. The height of the image is 0.287 m.

2. The wavelength of the light is 563 nm.

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Explanation:

The bonds that keep molecules together break apart and form new bonds during chemical reactions, rearranging atoms into different substances. Each bond takes a distinct amount of energy to either break or form; the reaction does not take place without this energy, and the reactants stay as they were.

The net force on particle q₁ is 6.76 x 10⁻³ N.

The net electric force acting on a charged particle due to other charged particles is given by Coulomb's law:

F = k × (|q₁|×|q₂|)/r²

where F is the force, k is Coulomb's constant, q₁ and q₂ are the charges on the particles, and r is the distance between them.

The direction of the force is along the line connecting the two charges, and is attractive for opposite charges and repulsive for like charges. In this case, the net force on q₁ due to q₂ and q₃ is the vector sum of the individual forces. So we need to calculate the force on q₁ due to q₂ and the force on q1 due to q₃ separately, and then add them as vectors.

F1 = k × (|q₁||q₂|)/r₁², where r1 = 0.220 m

F1 = (8.99 x 10⁹ Nm²/C²) x (8.99 uC) x (5.16 uC) / (0.220 m)²

F1 = 1.04 x 10⁻³N (attractive force)

F2 = k × (|q₁||q₃|)/r₂², where r₂ = 0.330 m

F2 = (8.99 x 10⁹ Nm²/C²) x (8.99 uC) x (89.9 uC) / (0.330 m)²

F2 = 7.80 x 10⁻³ N (repulsive force)

The net force on q₁ is the vector sum of F₁ and F₂, taking into account their directions:

F_net = F₁ - F₂ = 1.04 x 10⁻³ N - 7.80 x 10⁻³ N

F_net = -6.76 x 10⁻³ N (attractive force)

So the net force on q₁ is 6.76 x 10⁻³ N, in the direction towards q₂.

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Answer:

Sleeping

Explanation:

Driving in traffic is stressful because you may be late for wherever you are going,

Opening a birthday present is stressful because you don't know what is inside until you open it.

Watching a soap opera is stressful because you do not know what will happen next,

Sleeping is not stressful as long as you don't bring stressful topics to bed.

The density of the SF₆ gas at the given pressure and temperature is 2.96 g/l.

The given parameters:

The molecular mass of the SF₆ gas is calculated as follows;

M of SF₆ = 32 + (6 x 19) = 146 g/mol

The density of the SF₆ gas is calculated by applying ideal gas law as follows;

\(PV = nRT\\\\PV = \frac{m}{M} RT\\\\PM = \frac{m}{V} RT\\\\PM = \rho RT\\\\\rho = \frac{PM}{RT}\)

where;

\(\rho = \frac{0.5 \times 146}{0.0821 \times 300} \\\\\rho = 2.96 \ g/l\)

Thus, the density of the SF₆ gas at the given pressure and temperature is 2.96 g/l.

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Answer:

0.015 seconds

Explanation:

5/330=0.015

Answer:

0.015 seconds

Explanation:

divide 5 m and 330 m/s to get 0.015 seconds.

Answer:

✏︎A. Matter made up of only one type of atom

Explanation:

✏︎An element is a pure substance which cannot be broken down by chemical means, consisting of atoms which have identical numbers of protons in their atomic nuclei. The number of protons in the nucleus is the defining property of an element, and is referred to as the atomic number.

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Answer:

Acceleration,10

Explanation:

First 4kg is your mass,

Second 40N is your force

,so you need your acceleration

The answer is 10 because

F=40N

M=4kg

A=10

You divide 40 out of 4 and you get 10

HOPE THIS HELPS:)

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According to the data given in the question the net force on the box is of 5 N.

All of the forces that are applied to an object are added up to form the net force. As a consequence of the fact that it (force) is a vector and therefore that two forces with identical magnitudes and opposing directions cancel each other out, the resultant force is the total of the forces, or put another way, the net force is just the total of all the forces.

Given data :

Force on box to the left side (F1) = 15 N

Force on box to the right side (F2) = 20 N

Because both forces are in opposite direction

Hence,

Net force = F2 - F1

Net force = 20 - 15

Net force = 5 N.

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The centripetal force acting on a 1.5 kg mass moving in a circular path is 27N

A centripetal force is a net force acting on an object in order to maintain the object's movement in a circular motion.

According  to Newton's first law, it states an object will continue to proceed its movement in a straight line unless acted upon by an external force.

The centripetal force is the external force at work here and  It's the net force that propels the object in a circular motion.

Using the formula for calculating centripetal force:

\(\mathbf{F_c = mass (m) \times acceleration (a)}\)

\(\mathbf{F_c = 1.5 \ kg \times18 \ m/s^2}\)

\(\mathbf{F_c = 27 \ N}\)

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The electric field intensity at the same height but 2 meters away from the charge of a 1C electric charge is placed 1 meter above an infinite perfect conductor plane is -2kq/d² and and electric potential is -2kq/d.

The image method is a technique for calculating the electric field around a point charge placed near a conducting surface. The method involves creating an image charge on the opposite side of the conducting surface as the original point charge, which is a mirror of the original charge with respect to the surface. This image charge creates an electric field that cancels out the electric field created by the original charge at points on the surface.

To find the electric field intensity and electric potential at a point which is at a distance of 2 meters above the conducting plane and in line with the point charge, let’s assume that the image charge is located at a distance ‘d’ below the conducting plane. Therefore, the potential due to the image charge at a point P (which is at a distance of 2 meters above the conducting plane and in line with the point charge) will be,

Vi = -kq/d... (i)

where k is Coulomb’s constant and q is the charge of the point charge. As the image charge is on the opposite side of the conducting plane, the potential at the point P due to the image charge will be,

Vi’ = -kq/d... (ii)

Using the principle of superposition, the total potential at the point P is given as,

V = Vi + Vi’

V = -kq/d - kq/d

V = -2kq/d

Therefore, the electric field intensity at the point P due to the point charge will be,

E = -dV/dy

E = -d/dy(-2kq/d)

E = -2kq/d²

We have already calculated the potential due to the image charge at point P in equation (ii),

Vi’ = -kq/d

Therefore, the electric potential at point P due to the point charge is given as,

V = Vi + Vi’

V = -kq/d + (-kq/d)

V = -2kq/d

Therefore, the electric potential at the point which is 2 meters away from the charge and in line with it is given by, -2kq/d.

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The car's speed as it coasts into the gas station on the other side of the valley is 29.4 m/s.

At the top of  first hill,  total mechanical energy of the system is given by:

\(E = mgh + 1/2 mv^2\)

where m is the mass of  car, g is the acceleration due to gravity, h is  height of  hill, and v is the velocity.

Substituting values, we get:

\(E = (1250 kg)(9.81 m/s^2)(10 m) + 1/2 (1250 kg)(17 m/s)^2\)

\(E = 767125 J\)

At the bottom of the valley,  total mechanical energy of the system is given by:

\(E = mgh + 1/2 mv^2\)

Setting the initial mechanical energy equal to the final mechanical energy, we get:

\(mgh + 1/2 mv^2 = mgh' + 1/2 mv'^2\)

Car has run out of gas, we can assume that there is no external work done on the car.

Solving for v':

\(v' = sqrt(2gh + v^2 - 2gh')\)

Substituting  given values, we get:

\(v' = sqrt(2(9.81 m/s^2)(10 m) + (17 m/s)^2 - 2(9.81 m/s^2)(15 m))\)

\(v' = 29.4 m/s\)

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Answer: The color is orange, the state of matter is liquid

Explanation:

59.00 m is the calculated distance.

Distance is the sum of an object's movements, regardless of direction. Distance can be defined as the amount of space an object has covered, regardless of its starting or ending position. The International System of Units defines a meter as the distance unit. It's interesting to note that many other derived units or quantities, such as volume, area, acceleration, and speed, may be constructed using this as the basic unit and a few formulae.

V = V₀ + at = 4.30 + 3.00*5.00 = 19.30 m/s

d = V₀t + (1/2) at²

= 4.30*5.00 + (1/5.00) ² = 59.00 m

or

d = (1/2) (V₀+V) t

= (1/2) (4.30+19.30) (5.00)

= 59.00 m

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True: From the best current data, we infer the universe will expand forever.

As of my knowledge cutoff in September 2021, based on the best current data and observations, scientists infer that the universe will continue to expand forever. This inference is supported by the observations of the accelerated expansion of the universe, which was discovered in the late 1990s and confirmed by subsequent measurements.

This acceleration suggests the presence of dark energy, a hypothetical form of energy that permeates space and drives the expansion. However, scientific understanding and research are continuously evolving, so it's important to note that new observations or theories could potentially alter our current understanding in the future.

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Answer:

5.61s

Explanation:

\(\pink{\frak{Given}}\begin{cases} \textsf{ A ball is dropped off a tall building and accelerates at 9.8 m/s$^2$ .}\\\textsf{It reaches the ground at a speed of 55 m/s .} \end{cases}\)

And we need to find out the time taken by the ball to reach the ground .

Here we can use the First equation of motion , namely ;

\(\sf \longrightarrow v = u + at \)

where the symbols have their usual meaning . Now substituting the respective values , we have,

\(\sf \longrightarrow 55m/s = 0m/s + 9.8 m/s^2(t)\\ \)

\(\sf \longrightarrow 55m/s = 9.8m/s^2(t)\\ \)

\(\sf \longrightarrow t = \dfrac{55m/s}{9.8m/s^2}\\\)

\(\sf \longrightarrow \boxed{\bf time = 5.61 s }\)

The velocity to time graph of the robots can be used to calculate the displacement and distance traveled by the robot.

6) The magnitude of the displacement of the robot for the entire time interval shown is (ii) greater in Case B

7) The rank of the distance traveled during the intervals is as follows;

\(\begin{vmatrix}D| &A | &C | &B | &E| &F \\\end{vmatrix}\)

A velocity time graph describes the changes in the motion of a straight moving object

6) The displacement is given by the area under the velocity time graph.

The area is obtained by drawing a vertical line at the tip of the graph of the velocity to time of each of the robots

The area of a triangle = 0.5 × Base length × Height

The time interval in Case A is the same as the time interval in Case B which is taken as t

The time at which the velocity of the robot is 0 in figure (i) is approximately 0.5·t

The initial velocity in Case A = -v

The final velocity = v

The area under the curve in Case A, which gives the displacement is therefore;

Displacement in Case (A) = 0.5 × 0.5·t × (-v) + 0.5 × 0.5·t × (v) = 0

The magnitude of the initial velocity of the robot in Case B is half the magnitude of the final velocity of the robot

Similarly, the time at which the velocity of the robot in Case B is 0 is a third of the total time of motion

The displacement of the robot in Case B is therefore;

Displacement = 0.5 × (1/3)·t × (0.5·v) + 0.5 × (2/3)·t × (-v) = -0.25·v·t

The magnitude of the displacement in Case B = |-0.25·v·t| = 0.25·v·t > 0

The magnitude of the displacement in Case B is > The magnitude of the displacement in Case A

The correct option is therefore (ii) greater in Case B

7) The distance traveled is given by the area bounded by the curve as follows;

A. Shape of region bounded by the graph = Rectangle

Initial velocity = Constant = -1 unit

Time of travel = 5 unit

Area = -1 × 5 = -5 units

B. Shape of graph = Triangle

Initial velocity = 1.5 units

The final velocity = 0

Change in velocity = 1.5 - 0 = 1.5

The time = 5 unit

Area = 0.5 × 5 × 1.5 = 3.75

The distance = 3.75 units

C. Starting velocity = -1.7 units

The final velocity = 0

Change in velocity = -1.7

Time = 5 units

Distance ≈ 0.5 × 5 × -1.7 = -4.25

D. Shape of graph = Rectangle

Initial velocity ≈ 1.7 units (Constant)

Time = 5 units

Distance = 5 × 1.7 = 8.5

E. Shape of graph = Two congruent triangles

Initial velocity = -1 unit

Time it takes to increase to a velocity of 0 = 2.5·t

Final velocity = 1 unit

Time it takes to increase to from a velocity of 0 to a velocity of 1 = 2.5 units

Distance = Sum of the area of the two triangles

Displacement = 0.5 × 2.5 × -1 + 0.5 × 2.5 × 1 = 0

Distance = 0.5 × 2.5 × 1 + 0.5 × 2.5 × 1 = 2.5

F. Shape of area bounded by the graph = Positive triangle

Height of the triangle = 1 unit

Base length of the triangle = 5 units

Area of the triangle = Distance = 0.5 × 5 × 1 = 2.5

The rank of the distance travelled during the intervals is therefore;

D(8.5), A(-5), C(-4.25), B(3.75), E(2.5), F(2.5)

The distance is a measure of the motion of the robot such that the forward and backward motion are added together

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Answer:

I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.

Explanation:

I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.

I then cut the thread at this length and extend it out on a table or floor.

I then place the length of the thread along the meter rule and mark of the position of its tip and bottom along the meter rule.

The difference between the bottom and top points is the circumference of the cylinder.

If the thread is longer than the meter rule, i mark off where the the length of the meter rule stops along the thread and then place the meter rule at that point and then mark off where the length of the thread stops.

The circumference of the cylinder is then 1 meter plus the extra measurement along the meter rule.

Answer:

0.02

430

604

70,230

Explanation:

⇒zero in-between two digits or at the end of a whole is considered significant, example 604, 430, 70,230.

⇒zero at the end of a number with decimal point is not significant, example 7.560

⇒ zero before a digit in a number with decimal point is significant, example 0.02

Thus, the numbers which show zero as a significant digits include;

0.02

430

604

70,230

The kinetic energy of the spring is obtained as 12 J. Option C

We know that the kinetic energy is the energy that is required to stretch the spring. We can use the graph that has been shown to show the force constant of the spring.

We can find the force constant as the slope of the graph as follows;

K = \(y_{2} - y_{1} /x_{2} - x_{1}\)

K = 60 - 0/0.3 - 0.2

K = 60/0.1

K = 600 N/m

We then have;

W = KE = 1/2 Kx^2

Thus;

KE = 0.5 * 600 * (20 * 10^-2)^2

KE = 12 J

It would have a kinetic energy of 12 J.

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All of the following are characteristic of a hybrid electric vehicle (HEV), except Lower fuel economy.

A hybrid vehicle that combines an electric propulsion system with a traditional internal combustion engine (ICE) system is known as a hybrid electric vehicle (HEV) (hybrid vehicle drivetrain). The use of an electric powertrain aims to produce either higher performance or better fuel economy than a traditional car.

There are various HEV kinds, and each one differs in how much it functions as an electric vehicle (EV). While hybrid electric buses, boats, cars, and tractors are all available, hybrid electric cars are the most popular type of HEV.

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The magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

Gravitational potential energy is the energy which a body posses because of its position.

The gravitational potential energy of a body is given as,

= G = Fr²/Mm

Here, (m) is the mass of the body, (F) is the gravitational force and (r) is the height of the body.

The mass of baby is 4.20 kg and mass of father is 100 kg. The distance between them is 0.200 m. Put the values in the above formula as:

= 6.67 X 10⁻¹¹ =  F (0.200)² / 100 X 4.20

= F = 7 X 10⁻⁷ N

The magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29×1011 m.

Put the values in the above formula again as:

= 6.67 X 10⁻¹¹ =  F (6.29 X 10¹¹)² / 1.9 X 10²⁷ X 4.20

= F = 1.35 X 10⁻⁶ N

Thus, the magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

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The molecular field for bend distortion is h_bend = -K_bend ∇² n and the molecular field for twist distortion is h_twist = -K_twist ∇² n + K_twist q₀ (∇ × n).

In the case of bend distortion, the director (the preferred orientation of the liquid crystal molecules) deviates from the local normal direction. The bend energy density is given by:

E_bend = (K_bend/2) × (∇ · n)²

where K_bend is the bend elastic constant and ∇ · n represents the divergence of the director field.

To determine the molecular field h associated with the bend distortion, we can take the functional derivative of the bend energy with respect to the director field n:

h_bend = δE_bend/δn

Using the Euler-Lagrange equation, this can be written as:

h_bend = -K_bend ∇² n

So, the molecular field for bend distortion is h_bend = -K_bend ∇² n.

Twist distortion:

In the case of twist distortion, the director rotates in a helical manner. The twist energy density is given by:

E_twist = (K_twist/2) × (∇ × n - q₀)²

where K_twist is the twist elastic constant and q₀ represents the pitch of the helix.

Similarly, to determine the molecular field h associated with the twist distortion, we take the functional derivative of the twist energy with respect to the director field n:

h_twist = δE_twist/δn

Using the Euler-Lagrange equation, this can be written as:

h_twist = -K_twist ∇² n + K_twist q₀ (∇ × n)

So, the molecular field for twist distortion is h_twist = -K_twist ∇² n + K_twist q₀ (∇ × n).

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